Add one more solution for anagrams check (#4175)

src/main/java/com/thealgorithms/strings/Anagrams.java

@@ -43,6 +43,8 @@ public class Anagrams {
          * Auxiliary Space Complexity : O(1)
          * 4th approach Time Complexity : O(n)
          * Auxiliary Space Complexity : O(n)
+         * 5th approach Time Complexity: O(n)
+         * Auxiliary Space Complexity: O(1)
          */
     }
 
@@ -122,4 +124,27 @@ public class Anagrams {
             return nm.equals(kk);
         }
     }
+
+    boolean approach5(String s, String t) {
+        if(s.length() != t.length()){
+            return false;
+        }
+        // Approach is different from above 4 aproaches. 
+        // Here we initialize an array of size 26 where each element corresponds to the frequency of a character. 
+        int[] freq = new int[26];
+        // iterate through both strings, incrementing the frequency of each character in the first string and decrementing the frequency of each character in the second string. 
+        for(int i=0; i<s.length(); i++){
+            int pos1 = s.charAt(i) - 'a';
+			int pos2 = s.charAt(i) - 'a';
+            freq[pos1]++;
+            freq[pos2]--;
+        }
+        // iterate through the frequency array and check if all the elements are zero, if so return true else false
+        for(int i=0; i<26; i++){
+            if(freq[i] != 0){
+                return false;
+            }
+        }
+        return true;
+    }
 }

src/test/java/com/thealgorithms/strings/AnagramsTest.java

@@ -18,5 +18,6 @@ public class AnagramsTest {
         assertTrue(anagrams.approach3(input1, "teal"));
         assertTrue(anagrams.approach4(input1, "tale"));
         assertTrue(anagrams.approach4(input1, "teal"));
+        assertTrue(anagrams.approach5(input1, "teal"));
     }
 }